a) To calculate $S_3(5)$, we need to count the number of $(a, b, c, d, e)$ such that $a, b, c, d, e \in \{1, 2, 3\}$ and satisfy the conditions ii), iii). We investigate these two cases.

* If the first 3 terms are distinct, we consider the case where $(a, b, c) = (1, 2, 3)$ represents. For $(1, 2, 3, d, e)$ then $d=1$ or $d=2$, but if $d=1$ then there is no choice for $e$. Hence, $d=2$ and $e=1$. So this case has only one satisfying set. Note that there are $3! = 6$ ways to choose for $(a, b, c)$ so there are 6 satisfying tuples.

* If there are two identical numbers in the first 3 terms, we consider $(a, b, c) = (1, 2, 1)$ to represent. With set $(1, 2, 1, d, e)$, we see that $d$ cannot be 1 or 2 so $d=3$, thus $e=1$. Similarly, we can compute that there are exactly 6 satisfying sets.

Thus, $|S_3(5)| = 6 + 6 = 12$.

b) Call a set beautiful if it satisfies the given conditions. We will prove that the necessary and sufficient conditions for the existence of a nice set of numbers in the set $S_n(d)$ are $d \le 2n-1$.

Sufficient condition. For $d = 2n-1$, we consider a set of numbers of the form

$1, 2, 3, \dots, n-1, n, n-1, \dots, 3, 2, 1.$

By direct checking, we obtain that the upper set is beautiful. It follows that for every $1 \le d \le 2n - 1$, the set $S_n(d)$ is non-empty.

Necessary condition. To prove $S_n(d) = \emptyset$ when $d \ge 2n$, we only need to prove that there is no satisfying tuple when $d = 2n$. We will prove this by induction.

For $n = 1, d = 2$, the statement is clearly true. Assume that for every $1 \le k \le n$, there does not exist a beautiful tuple when $d = 2k$. We will show that this is also true for $k = n + 1$. Suppose that there exists a beautiful tuple for $d = 2k = 2(n + 1)$, which is $(x_1, x_2, \dots, x_{2n+2})$ then let $S$ be the number of the occurrences of the element $n+1$ in that tuple, we have the following cases:

* If $S = 0$ then this nice tuple has length $2n+2$ but the elements only take values not exceed $n$, which is a contradiction.

* If $S = 1$. If $n + 1$ is at the beginning or the end of the tuple, we delete it and the new set is also beautiful with the length is $2n + 1$, which is a contradiction. Therefore, the position of $n + 1$ must be centered with the form

$$(***, u, n+1, v, ***)$$

If $u \neq v$ then deleting $n+1$ again as above; and if $u = v$, removing $n+1$ along with $u$ or $v$ (so that no two equal numbers are next to each other), then the new set is also beautiful and has a length of $2n$, which is a contradiction.

* From the above argument, we observe that it is impossible for any value to appear exactly one time. Therefore, if $S = 2$ then all the values $1, 2, 3, \dots, n+1$ appear exactly two times in the tuple. Let $T$ be the number between two elements $n+1$ in the form

$$***, n+1, \underbrace{***}_{T \text{ numbers}}, n+1, ***$$

then it is clear that the outer numbers must be different from these $T$ numbers. Without loss of generality, suppose that these number takes the value of $\{1, 2, \dots, m\}$, then $T = 2m < 2n$ and these numbers form a new beautiful tuple, which is a contradiction.

* If $S \ge 3$, by the above argument, this case does not happen.

Therefore, in all cases, the assertion is true. Thus, $|S_n(d)| > 0$ if and only if $d \le 2n - 1$. $\square$